I know that if $f$ is a uniform continuous, then $f(x_n)$ is Cauchy if $x_n$ is Cauchy.
Assume $x_n$ is a Cauchy sequence, and $f$ is a discontinuous function only at certain finite points.
I am trying to prove $f(x_n)$ "kills" the Cauchy quality.
I know the format of the proof for any Cauchy function is to show that $\forall \epsilon > 0$, $\exists N \in \mathbb N$ such that if $n,m\ge N$, then $|x_n - x_m|$ < $\epsilon$.
Can I just set the restrictions that $n,m \neq ({x_j,x_{j+1}}...)$ which describes the set of discontinuous points? The only proof I need to show is that there exists SOME $n,m \ge N$. Outside of these point, $f$ is uniform continuous.
Then, $\forall \epsilon>0$, $f$ is uniformly continuous for these special points, $\exists\delta>0$ such that $|f(a)-f(b)|<\epsilon$ for $|a-b|<\delta$.
Since $x_n$ is Cauchy, there exists $N>0$ such that $|x_n-x_m|<\delta$ for $m,n>N$
Hence$|f(x_n)-f(x_m)|<\epsilon$ for $m,n>N$. So $f(x_n)$ is Cauchy, and the Cauchy property is not killed. Again, this proof was constructed by assuming there were $n,m$ values $\ge N \in \mathbb N$.
It is no way for you to show in general that $f$ kills the Cauchy property: For example, if $f$ is given by
$$f(x) = \begin{cases} 1 & \text{if } x \neq 0\\ 0 & \text{if } x = 0 .\end{cases}$$
Then $f$ is not continuous at $0$. However, for every $\{x_n\}$ Cauchy sequence which does not contain $0$, $f(x_n)$ is Cauchy: It's a constant sequence.
Even if $f$ has no limit as $x \to 0$, then still there might be some Cauchy sequence $x_n\to 0$ such that $f(x_n)$ is Cauchy: $f(x) = \sin \frac 1x$ and $x_n = \frac 1{2\pi n}$ is an example.
I think the best you have is the following: If $f$ is discontinuous at $a$, then there is a Cauchy sequence $x_n \to a$ so that $f(x_n)$ is not Cauchy. To see this, note that if $f$ is continuous at $a$ if and only if for all sequence $x_n \to a$, then we have $f(x_n) \to f(a)$. So if $f$ is not continuous at $a$, there is a sequence $x_n \to a$ so that $f(x_n)$ does not converge to $f(a)$. There are two cases: If $f(x_n)$ does not converges at all, then it's not Cauchy and you are done. If $f(x_n)$ does converge to $b \neq f(a)$, then setting the new sequence
$$\{ x_1, a, x_2, a, x_3, a, \cdots \},$$
this new sequence still converges to $a$, but $f$ of this sequence is not Cauchy as it is not convergent.